GATE Breaker

The literal count of a Boolean expression is the sum of the number of times each literal appears in the expression. For example, the literal count of (xy + xz′) is 4. What are the minimum possible literal counts of the product-of-sum and sum-of-product representations, respectively, of the function given by the following Karnaugh map? Here, X denotes “don’t care”.  (A) (11, 9)  (B) (9, 13)  (C) (9, 10)  (D) (11, 11)  (GATE 2003)  Answer: (C) (9, 10)  Explanation:  Sum-of-product (SOP) = WY + W'Y' + Z'WX' + XYZ'  Therefore, literal count = 10  Product-of-sum (POS) = (Y' + Z')(W' + Z')(Z' + Y)(X + Z + W)   Therefore, linteral count = 9  Previous Post Next Post

The IEEE standard for floating-point numbers specifies how the single-precision (32-bit) and double-precision (64-bit) floating-point numbers are represented.  1. Single Precision:   The IEEE single-precision floating-point standard representation requires a 32-bit word. The first bit to the left is the sign bit , the next eight bits are the exponent bits , and the final 23 bits form the fraction part .  The value of a normalized number $= (-1)^s \times 1.m \times 2^{e^{-127}}$, where $s = 1$ fornegative number and $s = 0$ for the positive number, $m = $ mantisa and $e = $ exponent.  2. Double Precision:   The IEEE double-precision floating-point standard representation requires a 64-bit word. The first bit is the sign bit , the next eleven bits are the exponent bits , and the final 52 bits form the fraction part .   Example Problem:   The following is a scheme for floating-point number representation using 16 bits.  Bit Position 15 14...9 8...0 ...

In this tutorial, we will learn how to compute the difference between two binary numbers using 1's complement and 2's complement.  Let us consider two binary numbers $X$ and $Y$.  We want to compute $X - Y$ here.  Method 1: Using 1's complement  Step 1: Find 1's complement of $Y$.  Step 2: Add $X$ and $Y$.  Step 3: If there is a carry, then add this carry to the answer. If there is no carry, then find the 1's complement of the answer and place a negative sign in front of the answer.  Example 1: Compute $(101)_2 - (011)_2$ using 1's complement method.   Solution:  Here, $X = (101)_2$ and $Y = (011)_2$.  Step 1: 1's complement of $Y$ is $(100)_2$.  Step 2: $X + Y = (101)_2 + (100)_2 = (1 001)_2$.  Step 3: There is a carry of 1, so now we will add this carry to the answer.  Therefore, the final result is $(1)_2 + (001)_2 = (010)_2$.  Example 2: Compute $(011)_2 - (101)_2$ using 1's complement method.  Solution...

In a game, two players $X$ and $Y$ toss a coin alternately. Whosoever gets a 'head' first wins the game and the game is terminated. Assuming that player $X$ starts the game. The probability of player $X$ winning the game is  (A) $\dfrac{1}{3}$  (B) $\dfrac{1}{2}$  (C) $\dfrac{2}{3}$  (D) $\dfrac{2}{4}$  Answer: (C) $\dfrac{2}{3}$  Explanation:  Here, the probability of getting a 'head' is, $p = \dfrac{1}{2}$, and  the probability of getting a 'tail' is, $q = \dfrac{1}{2}$.  Therefore, the required probability is  $p + q^2p + q^4p + \cdots$  $= p(1 + q^2 + q^4 + \cdots)$  $= p \times \dfrac{1}{1 - q^2}$  $= \dfrac{1}{2} \times \dfrac{1}{1 - \dfrac{1}{4}}$  $= \dfrac{2}{3}$ 

For a random variable $X$ ($-\infty < X < \infty$) following normal distribution, the mean is $\mu = 100$. If the $P(X \geq 110)$, the find $P(90 \leq X \leq 110)$.  (A) $1 - 2\alpha$  (B) $1 - \alpha$  (C) $1 - \dfrac{\alpha}{2}$  (D) $2\alpha$  Answer: (A) $1 - 2\alpha$  Explanation:  Here, $X$ is a random variable that follows a normal distribution with mean $\mu = 100$.  Given, $P(X \geq 110) = \alpha$  Hence, $P(X \leq 90) = \alpha$  Therefore, $P(90 \leq X \leq 110) = 1 - 2\alpha$. 

A coin is tossed 4 times. What is the probability of getting heads exactly 3 times?  (A) $\dfrac{1}{4}$  (B) $\dfrac{3}{8}$  (C) $\dfrac{1}{2}$  (D) $\dfrac{3}{4}$  Answer: (A) $\dfrac{1}{4}$  Explanation:  Let $X$ be a random variable that denotes the number of heads.  Here, $X$ follows the binomial distribution with the parameters $p = \dfrac{1}{2}$ and $n = 4$.  Hence, $P(X = r) = ^nC_r \times p^r \times (1 - p)^{n - r}$, where $r = 1, \ldots, n$.  Therefore, the required probability is  $P(X = 3) = ^4C_3 \times (\dfrac{1}{2})^3 \times (1 - \dfrac{1}{2})^{4 - 3}$  $= 4 \times \dfrac{1}{8} \times \dfrac{1}{2}$  $= \dfrac{1}{4}$ 

A random variable is uniformly distributed over the interval 2 to 10. Its variance will be  (A) $\dfrac{16}{3}$  (B) 6  (C) $\dfrac{256}{9}$  (D) 36  Answer: (A) $\dfrac{16}{3}$  Explanation:  The expected value (i.e. the mean) of a uniform random variable $X$ is:  $E(X) = \dfrac{b + a}{2}$.  The variance of a uniform random variable $X$ is:  $Var(X) = \dfrac{(b - a)^2}{12}$.  Here, $a$ is the minimum value in the distribution, and $b$ is the maximum value.  For the given problem, $a = 2$ and $b = 12$.  Therefore, $Var(X) = \dfrac{(10 - 2)^2}{12} = \dfrac{16}{3}$. 

$X$ is a uniformly distributed random variable that takes values between zero and one. The value of $E( X^3)$ will be  (A) 0  (B) $\dfrac{1}{8}$  (C) $\dfrac{1}{4}$  (D) $\dfrac{1}{2}$  Answer: (C) $\dfrac{1}{4}$  Explanation:  Here, $X$ is a uniformly distributed random variable that takes values between 0 and 1.  Therefore, its p.d.f. is $f(x) = \dfrac{1}{1 - 0} = 1, \mbox{for } 0 < x < 1.$  Hence,  $E(X^3) = \int\limits_{0}^{1}x^3f(x)dx = \int\limits_{0}^{1}x^3dx = \dfrac{1}{4}$  Previous Post Next Post

An examination consists of two papers, Paper-1 and Paper-2. The probability of failing in Paper-1 is 0.3 and that in Paper-2 is 0.2. Given that a student has failed in Paper-2, the probability of failing in Paper-1 is 0.6. The probability of a student failing in both the papers is  (A) 0.5  (B) 0.18  (C) 0.12  (D) 0.06  Answer: (C) 0.12  Explanation:  Let $E_1$ and $E_2$ be the two events of failing in Paper-1 and Paper-2 respectively.  Given, $P(E_1) = 0.3$, $P(E_2) = 0.2$ and $P(E_1 | E_2) = 0.6$  Now, $P(E_1 | E_2) = \dfrac{P(E_1 \cap E_2)}{P(E_2)}$  $\Rightarrow 0.6 = \dfrac{P(E_1 \cap E_2)}{0.2}$  $\Rightarrow P(E_1 \cap E_2) = 0.12$, which is the required probability. 

If X is a continuous random variable whose probability density function is given by \[f(x) = \left\{ \begin{array}{ll} k(5x - 2x^2) & \mbox{if } 0 \leq x \leq 2 \\ 0 & \mbox{otherwise,} \end{array} \right.\] then $P(X > 1)$ is  (A) $\dfrac{3}{14}$  (B) $\dfrac{4}{5}$  (C) $\dfrac{14}{7}$  (D) $\dfrac{17}{28}$  Answer: (D) $\dfrac{17}{28}$  Explanation:  $\int\limits_{-\infty}^{+\infty}f(x)dx = 1$  $\Rightarrow \int\limits_{0}^{2}k(5x-2x^2)dx = 1$  $\Rightarrow k = \dfrac{3}{14}$  Now, $P(X > 1)$  $= \int\limits_{1}^{\infty}f(x)dx$  $= \int\limits_{1}^{2} \dfrac{3}{14}(5x - 2x^2)dx$  $= \dfrac{17}{28}$